The points $\left(\mathrm{k},2-2\mathrm{k}\right), (1-\mathrm{k},2\mathrm{k})$ and $(-4-\mathrm{k},6-2\mathrm{k})$ are collinear. Then $\mathrm{k}=$

Let A=(k, 2-2k), B=(1-k, 2k), C=(-4-k, 6-2k) are vertices of triangle ABC

Since given points are collinear then area of triangle=0

$$\begin{gathered} \Rightarrow \frac{1}{2}\left|\begin{array}{cc}{\mathrm{x}}_1-{\mathrm{x}}_2& {\mathrm{y}}_1-{\mathrm{y}}_2\\ {\mathrm{x}}_1-{\mathrm{x}}_3& {\mathrm{y}}_1-{\mathrm{y}}_3\end{array}\right|=0 \\ \Rightarrow \frac{1}{2}\overline{)\begin{array}{c}2\mathrm{k}-1\\ 2\mathrm{k}+4\end{array}\begin{array}{c}\mathrm{X}\end{array}}\overline{)\begin{array}{c}2-4\mathrm{k}\\ -4\end{array}}=0 \\ \Rightarrow \left|-4(2\mathrm{k}-1)-(2-4\mathrm{k})(2\mathrm{k}+4)\right|=0 \\ \Rightarrow \left|-8\mathrm{k}+4-(4\mathrm{k}+8-8{\mathrm{k}}^2-16\mathrm{k})\right|=0 \\ \Rightarrow \left|-8\mathrm{k}+4+12\mathrm{k}-8+8{\mathrm{k}}^2\right|=0 \\ \Rightarrow 8{\mathrm{k}}^2+4\mathrm{k}-4=0 \\ \Rightarrow 2{\mathrm{k}}^2+\mathrm{k}-1=0 \\ \Rightarrow 2{\mathrm{k}}^2+2\mathrm{k}-\mathrm{k}-1=0 \\ \Rightarrow 2\mathrm{k}(\mathrm{k}+1)-1(\mathrm{k}+1)=0 \\ \Rightarrow \mathrm{k}=-1 \left(\mathrm{or}\right) \mathrm{k}=\frac{1}{2} \end{gathered}$$